Let $f(x)=\csc(x)$. Find $f'\left(\dfrac{\pi}{2}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $\dfrac{1}{2}$ (Choice D) D $1$
Solution: Let's first find $f'(x)$. Then, we can evaluate it at $x=\dfrac{\pi}{2}$. Recall that the derivative of $\csc(x)$ is $-\dfrac{\cos(x)}{\sin^2(x)}$, or $-\csc(x)\cot(x)$. [Is there a way to know this without memorizing?] So $f'(x)=-\dfrac{\cos(x)}{\sin^2(x)}$. Now let's plug $x={\dfrac{\pi}{2}}$ into $f'$ : $\begin{aligned} &\phantom{=}f'\left({\dfrac{\pi}{2}}\right) \\\\ &=-\dfrac{\cos\left({\dfrac{\pi}{2}}\right)}{\sin^2\left({\dfrac{\pi}{2}}\right)} \\\\ &=-\dfrac{0}{\left(1\right)^2} \\\\ &=0 \end{aligned}$ In conclusion, $f'\left(\dfrac{\pi}{2}\right)=0$.